Polynomials and Their Properties: MCQ test, Set 7

Download the PDF (with key sheet and solutions, 91.4 KB)

Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

Pick the monomial from the following.
A $\displaystyle -\frac{5}{x^{2}}$ B $\displaystyle -\frac{6}{x^{3}}$ C $\displaystyle -8x^{\frac{1}{3}}y^{4}$ D $\displaystyle -\frac{4}{3}x^{2}$

Answer and solution

Answer: (D) $\displaystyle -\frac{4}{3}x^{2}$

$\displaystyle -\frac{4}{3}x^{2}$ : the powers of $x$ are $2$ , non-negative integers, so it is a monomial.

Each of the others has a negative or fractional power.
The degree of the monomial $5x^{5}y^{4}z^{4}$ is
A $14$ B $5$ C $13$ D $80$

Answer and solution

Answer: (C) $13$

Degree = sum of the exponents $= 5 + 4 + 4 = 13$
The coefficient of $x^{3}$ in $-4 + 2x^{2} + 8x - 9x^{3}$ is
A $2$ B $-9$ C $3$ D $9$

Answer and solution

Answer: (B) $-9$

The term with $x^{3}$ is $-9x^{3}$ , so the coefficient is $-9$ .
$6 + 4x$ is a
A Bi-quadratic polynomial B Quadratic polynomial C Cubic polynomial D Linear polynomial

Answer and solution

Answer: (D) Linear polynomial

The highest power of $x$ is $1$ , so it is a linear polynomial.
Pick the polynomial from the following.
A $-4\sqrt{5}x^{5} - 3x$ B $3\sqrt{2}x^{3} - 2\sqrt{x}$ C $\displaystyle 2x^{2} - 6\sqrt{x} + \frac{1}{3}$ D $\displaystyle -\frac{7}{3}x^{3} - 5x^{\frac{1}{3}}$

Answer and solution

Answer: (A) $-4\sqrt{5}x^{5} - 3x$

In $-4\sqrt{5}x^{5} - 3x$ , all powers of $x$ are whole numbers, so it is a polynomial.

Each other option has a negative or fractional power of $x$ .
If $ax^{n-6} - bx^{n-7} - cx^{n-8}$ is a cubic polynomial ( $n \in N$ ), then $n =$
A $11$ B $10$ C $9$ D $3$

Answer and solution

Answer: (C) $9$

The degree of a cubic polynomial is $3$ .

So the highest power: $n-6 = 3$

$\Rightarrow n = 9$
The equation $-6x + 4 = 0$ is a
A Quadratic equation B Bi-quadratic equation C Linear equation D Cubic equation

Answer and solution

Answer: (C) Linear equation

The highest power of $x$ is $1$ , so it is an linear equation.
If $p(x) = -x^{3} - 4x^{2} - 2x + 6$ , then $p(-2) =$
A $-6$ B $2$ C $34$ D $-22$

Answer and solution

Answer: (B) $2$

$p(-2) = -(-2)^{3} - 4(-2)^{2} - 2(-2) + 6$

$= 8 - 16 + 4 + 6$

$= 2$
The zero of the polynomial $7x + 5$ is
A $\displaystyle -\frac{7}{5}$ B $\displaystyle -\frac{5}{7}$ C $\displaystyle \frac{7}{5}$ D $\displaystyle \frac{5}{7}$

Answer and solution

Answer: (B) $\displaystyle -\frac{5}{7}$

$\displaystyle 7x + 5 = 0 \Rightarrow 7x = -5 \Rightarrow x = -\frac{5}{7}$
If $-1$ is a zero of $p(x) = x^{3} - x^{2} + ax - 2$ , then $a =$
A $-3$ B $4$ C $2$ D $-4$

Answer and solution

Answer: (D) $-4$

$p(-1) = (-1)^{3} - (-1)^{2} + a(-1) - 2$

$= -1 - 1 - a - 2$

$= -a - 4$

$-a - 4 = 0 \Rightarrow a = -4$
The roots of $6x^{2} - 17x + 7 = 0$ are
A $\displaystyle -\frac{1}{2}, -\frac{7}{3}$ B $\displaystyle \frac{1}{2}, \frac{7}{3}$ C $\displaystyle 2, \frac{3}{7}$ D $1, 7$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{2}, \frac{7}{3}$

$6x^{2} - 17x + 7 = 6x^{2} - 3x - 14x + 7$

$= 3x(2x - 1) - 7(2x - 1)$

$= (2x - 1)(3x - 7)$

$\displaystyle \Rightarrow x = \frac{1}{2}, \frac{7}{3}$
The roots of $2x^{2} + 6x + 1 = 0$ are
A $\displaystyle \frac{3 \pm \sqrt{7}}{2}$ B $\displaystyle \frac{-3 \pm \sqrt{7}}{2}$ C $-3 \pm \sqrt{7}$ D $\displaystyle \frac{-3 \pm \sqrt{11}}{2}$

Answer and solution

Answer: (B) $\displaystyle \frac{-3 \pm \sqrt{7}}{2}$

$a = 2, b = 6, c = 1$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 4(2)(1)}}{2(2)}$

$\displaystyle = \frac{-3 + \sqrt{7}}{2}, \frac{-3 - \sqrt{7}}{2}$
The remainder when $2x^{4} - 5x^{3} - 11x^{2} - 11x - 7$ is divided by $x^{2} - 4x - 1$ is
A $4x - 4$ B $-4x + 4$ C $4x + 4$ D $-4x - 4$

Answer and solution

Answer: (A) $4x - 4$

$2x^{4} \div x^{2} = 2x^{2}$ ; $2x^{2} \times (x^{2} - 4x - 1) = 2x^{4} - 8x^{3} - 2x^{2}$ ; subtract: $3x^{3} - 9x^{2} - 11x - 7$

$3x^{3} \div x^{2} = 3x$ ; $3x \times (x^{2} - 4x - 1) = 3x^{3} - 12x^{2} - 3x$ ; subtract: $3x^{2} - 8x - 7$

$3x^{2} \div x^{2} = 3$ ; $3 \times (x^{2} - 4x - 1) = 3x^{2} - 12x - 3$ ; subtract: $4x - 4$

Remainder $= 4x - 4$
The remainder when $24x^{3} - 8x^{2} - 8x + 5$ is divided by $2x - 1$ is
A $2$ B $-2$ C $4$ D $13$

Answer and solution

Answer: (A) $2$

$\displaystyle 2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ ; remainder $\displaystyle = f\left(\frac{1}{2}\right)$

$= 2$
If $x + 1$ is a factor of $x^{3} - 3x^{2} + x + a$ , then $a =$
A $4$ B $-5$ C $1$ D $5$

Answer and solution

Answer: (D) $5$

$f(-1) = (-1)^{3} - 3(-1)^{2} + (-1) + a$

$= -1 - 3 - 1 + a$

$= a - 5$

$a - 5 = 0 \Rightarrow a = 5$
$5x^{2} + 46x + 9 =$
A $(x + 1)(5x + 9)$ B $(x - 1)(5x - 9)$ C $(5x + 1)(x + 9)$ D $(x - 9)(5x - 1)$

Answer and solution

Answer: (C) $(5x + 1)(x + 9)$

$5x^{2} + 46x + 9 = 5x^{2} + x + 45x + 9$

$= x(5x + 1) + 9(5x + 1)$

$= (5x + 1)(x + 9)$
What must be added to $x^{3} + 2x^{2} + 9x + 5$ so that it is exactly divisible by $x + 3$ ?
A $-31$ B $77$ C $31$ D $-77$

Answer and solution

Answer: (C) $31$

Remainder $= f(-3)$ (remainder theorem)

$f(-3) = (-3)^{3} + 2(-3)^{2} + 9(-3) + 5$

$= -27 + 18 - 27 + 5$

$= -31$

The remainder is $-31$ , so $31$ must be added.
If $x - 4$ and $x - 2$ are factors of $x^3 + ax^2 + bx - 24$ , then
A $a = -9, b = 26$ B $a = 9, b = -26$ C $a = -9, b = -26$ D $a = 9, b = 26$

Answer and solution

Answer: (A) $a = -9, b = 26$

$f(4) = 0 \Rightarrow 4a + b = -10$ ... (1)

$f(2) = 0 \Rightarrow 2a + b = 8$ ... (2)

$(1) - (2)$ : $2a = -18$

$\Rightarrow a = -9$

Put $a = -9$ in (1): $-36 + b = -10 \Rightarrow b = 26$
A cubic polynomial has zeros $1$ , $3$ and $-1$ and leading coefficient $3$ . The polynomial is
A $x^{3} - 3x^{2} - x + 3$ B $3x^{3} + 9x^{2} - 3x - 9$ C $x^{3} + 3x^{2} - x - 3$ D $3x^{3} - 9x^{2} - 3x + 9$

Answer and solution

Answer: (D) $3x^{3} - 9x^{2} - 3x + 9$

With zeros $1, 3, -1$ and leading coefficient $3$ : $3(x - 1)(x - 3)(x + 1)$

$= 3x^{3} - 9x^{2} - 3x + 9$
Given that $x - 2$ is a factor, $x^{3} - 12x^{2} + 44x - 48 =$
A $(x - 2)(x^{2} - 10x + 24)$ B $(x - 2)(x^{2} - 14x + 72)$ C $(x - 2)(x^{2} - 10x - 24)$ D $(x - 2)(x^{2} + 10x + 24)$

Answer and solution

Answer: (A) $(x - 2)(x^{2} - 10x + 24)$

$x^{3} \div x = x^{2}$ ; $x^{2} \times (x - 2) = x^{3} - 2x^{2}$ ; subtract: $-10x^{2} + 44x - 48$

$-10x^{2} \div x = -10x$ ; $(-10x) \times (x - 2) = -10x^{2} + 20x$ ; subtract: $24x - 48$

$24x \div x = 24$ ; $24 \times (x - 2) = 24x - 48$ ; subtract: $0$

Quotient $= x^{2} - 10x + 24$
$x^{3} - 6x^{2} + 5x + 12 =$
A $(x + 3)(x + 4)(x - 1)$ B $(x - 3)(x - 4)(x + 1)$ C $(x + 3)(x - 4)(x + 1)$ D $(x - 3)(x + 4)(x + 1)$

Answer and solution

Answer: (B) $(x - 3)(x - 4)(x + 1)$

$f(-1) = -1 - 6 - 5 + 12 = 0$

Dividing and splitting the quotient gives $f(x) = (x - 3)(x - 4)(x + 1)$ .
Using synthetic division, the quotient when $5x^{3} - 2x - 8$ is divided by $x - 3$ is
A $5x^{3} + 15x^{2} + 43x + 121$ B $5x + 13$ C $5x^{2} + 15x + 43$ D $5x^{2} - 15x + 43$

Answer and solution

Answer: (C) $5x^{2} + 15x + 43$

Second row: $0 \quad 15 \quad 45 \quad 129$

Bottom row: $5 \quad 15 \quad 43 \;\big|\; 121$

Quotient $= 5x^{2} + 15x + 43$
$27x^{3} + 1 = (3x + 1) \times$
A $9x^{2} - 3x + 2$ B $x^{2} - x + 1$ C $9x^{2} - 3x + 1$ D $9x^{2} + 3x + 1$

Answer and solution

Answer: (C) $9x^{2} - 3x + 1$

$27x^{3} + 1 = (3x + 1)(9x^{2} - 3x + 1)$

Signs alternate; the powers of $1$ go up as the powers of $3x$ go down.
When $\displaystyle \frac{27}{8}a^{3} + \frac{81}{4}a^{2}x + \frac{81}{2}ax^{2} + 27x^{3}$ is divided by $\displaystyle \frac{3}{2}a + 3x$ , the quotient is
A $\displaystyle \frac{9}{4}a^{2} - 9ax + 9x^{2}$ B $\displaystyle \frac{9}{4}a^{2} + 9x^{2}$ C $\displaystyle \frac{9}{4}a^{2} + 9ax - 9x^{2}$ D $\displaystyle \frac{9}{4}a^{2} + 9ax + 9x^{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{9}{4}a^{2} + 9ax + 9x^{2}$

$\displaystyle \frac{27}{8}a^{3} \div \frac{3}{2}a = \frac{9}{4}a^{2}$ ; $\displaystyle \frac{9}{4}a^{2} \times (\frac{3}{2}a + 3x) = \frac{27}{8}a^{3} + \frac{27}{4}a^{2}x$ ; subtract: $\displaystyle \frac{27}{2}a^{2}x + \frac{81}{2}ax^{2} + 27x^{3}$

$\displaystyle \frac{27}{2}a^{2}x \div \frac{3}{2}a = 9ax$ ; $\displaystyle 9ax \times (\frac{3}{2}a + 3x) = \frac{27}{2}a^{2}x + 27ax^{2}$ ; subtract: $\displaystyle \frac{27}{2}ax^{2} + 27x^{3}$

$\displaystyle \frac{27}{2}ax^{2} \div \frac{3}{2}a = 9x^{2}$ ; $\displaystyle 9x^{2} \times (\frac{3}{2}a + 3x) = \frac{27}{2}ax^{2} + 27x^{3}$ ; subtract: $0$
$x^2 + bx + c$ leaves the remainders $2$ and $-1$ when divided by $x - 1$ and $x - 2$ respectively. Then
A $b = -6, c = 7$ B $b = 6, c = 7$ C $b = -6, c = -7$ D $b = 6, c = -7$

Answer and solution

Answer: (A) $b = -6, c = 7$

$f(1) = 1 + b + c = 2 \Rightarrow b + c = 1$ ... (1)

$f(2) = 4 + 2b + c = -1 \Rightarrow 2b + c = -5$ ... (2)

$(1) - (2)$ : $-b = 6$

$\Rightarrow b = -6$

Put $b = -6$ in (1): $-6 + c = 1 \Rightarrow c = 7$

Menu

MCQ test, Set 7